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2 years ago

A semi has a net force of 13,000 N and is accelerating at a rate of 6.5 m/s^2. What is the mass of the semi? *

1 answer:

4 0

Example 1: An 850-kg car is accelerating at a rate of 2.4m/s2 to the right along a ... 2) A nonzero net force ΣF acting on mass M causes an acceleration a in it such that ΣF = Ma. The acceleration has the same direction as the applied net force. ... (b) Knowing that the crate is being pushed to the left by a 53-N force

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True or false? the electric field produced by a uniform, infinite plane of charge does not depend on the distance from the plane

wolverine [178]

True, the electric field produced by a uniform, infinite plane of charge does not depend on the distance from the plane.

<h3>What is electric field?</h3>

Electric field is the region of space at which electric force is felt.

E = kq/r²

where;

r is the distance of the charge

at infinite plane, r → ∞, and E = 0

Thus, the electric field produced by a uniform, infinite plane of charge does not depend on the distance from the plane.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

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1 year ago

One characteristic of a lagoon is often

pishuonlain [190]

Separated by a larger body of water

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3 years ago

Which scientist determined that the further a planet is from the sun, the longer it takes to orbit the sun? ~ I think it might b

lozanna [386]

That was Johannes Kepler.

He couldn't say why that should happen. But he was able to show that
if we assume the the sun is in the center of the solar system, and assume
that the planets revolve around the sun in elliptical orbits, and assume that
the farther a planet is from the sun the longer it takes to orbit the sun, then
that matches everything that we see the planets actually doing in the sky.

Then Isaac Newton came along, and wrote down a short, simple formula
for the force of gravity. And then he showed that IF his formula for gravity
is correct, then the planets MUST do everything that we see them actually
doing in the sky, and they MUST behave according to Kepler's laws.

Wotta guy !

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3 years ago

How to calculate t in shm when you have displacement.

sukhopar [10]

Answer:

ikd

Explanation:

I don't get what your asking pls make it more clear

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2 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kV and enter a uniform magnetic field of

3241004551 [841]

Answer:

see answer

Explanation:

First, we define the variables we are going to use

let:

$\Delta V = 2000 [V]$ the voltage difference

$q = 1.6*10^{-19} [C]$ the charge

$m = 238*1.6*10^{-27} [Kg]$ be the mass of U238, it comes from multiplying the isotopes mass in amu's by the mass of a proton

$B = 1.20 [T]$

We start by calculating the velocity of the ions, work is equal to charge times potential difference

$W = q\Delta V$ , which is nothing else than energy (the units are the same, Joules)

Therefore the kinetic energy of the particle will be equal to the magnitude of $q\Delta V$

let's find the velocity now :

$\frac{1m*v^2}{2}=q\Delta V => v = \sqrt{\frac{2q \Delta V}{m} }$

we get $v =40996 [m/s]$

a) We note that the magnetic field is perpendicular to the velocity of the ions, therefore we can use the lorentz force formula

$F = qvBsin( \theta)$ , where theta is the angle between B and V (which is 90 as they are perpendicular to each other), F is the centripetal force $\frac{mv^2}{r}$

so we get $qvBsin( 90) = \frac{mv^2}{r} => qB = \frac{mv}{r}$

we just replace v! => $r = \frac{mv}{qB} => r = \frac{m*(\sqrt{\frac{2q \Delta V}{m} })}{qB}$

m enters the square root and cancels out the inner m, and so does q.

we get now an expression for the radius of the circular path:

$r = \frac{1}{B} *(\sqrt{\frac{2m \Delta V}{q} })$

replacing values we obtain r = $\frac{1}{1.6}* \sqrt{\frac{2*238*1.6*10^-{27}*2000}{1.6*10^-19} } = 0.0609 [m]$

b) for uranium 235 we just replace the 238 with 235:

r = $\frac{1}{1.6}* \sqrt{\frac{2*235*1.6*10^-{27}*2000}{1.6*10^-19} } = 0.0605 [m]$

c) it depends heavily on the accelerating voltage, we just saw that the radius of the path does not change a lot with a change in the mass (0.0605 vs. 0.0609), if we drop the voltage difference to 1.00Kv we get for the 238 uranium ions a radius of 0.043, for the 235 uranium 0.042 which is approximately equal, the main factor is the voltage difference

d) it also depends heavily on the magnitude of the magnetic field, if we increase 1.6 T to let's say 2T we get a radiius of 0.034 for U-238. The decissive factor here is the angle theta between V and B, we got lucky here because they are perpendicular, in any other case smaller angles will decrease the magnitude of B.

3 0

3 years ago